Question

Show that $\large\tt { d(x,y) = \frac{ \left| x-y \right| }{ 1+ \left| x-y \right| }}$ defines as a metric on (0, $\large \infty$ ​

Answer 1

Answer:

$\large\tt { d_{2} (x,y) + d_{2} (y,z) = \frac {d (x,y)}{1+ d (x,y)} + \frac {d(y,z)}{1+d(y,z)}}$

$\large\tt { \geq \frac {d (x,y)}{1+d(x,y) + d(y,z)} + \frac {d (y,z)}{1+d(x,y) + d(y,z)} = \frac {d (x,y) d (y,z)} = {1+d(x,y) + d(y,z)}}$

= $\large\tt { 1– \frac {1}{1+ d(x,y) + d(y,z)} \geq 1- \frac {1}{1+d(x,z)}}$

$\large\tt { = d_{2} (x,z)}$

Answer 2

d2(x,y)+d2(y,z)

=d(x,y)1+d(x,y)+d(y,z)1+d(y,z)

≥d(x,y)1+d(x,y)+d(y,z)+d(y,z)1+d(x,y)+d(y,z)

=d(x,y)+d(y,z)1+d(x,y)+d(y,z)

=1−11+d(x,y)+d(y,z)≥1−11+d(x,z)

=d(x,z)1+d(x,z)

=d2(x,z)